# Measuring the natural frequency of a spring-mass system with the graph

On a graph of a system under a external force y = distance and x = time where the external force start at t = 0, it's easy to find the driving frequency.

$$F = \frac{\omega}{2\pi}, \omega = \frac{2\pi}{T}$$ and we can get $$T$$ easily with the steady state part of the graph.

However, is there a way to find the natural frequency?

Maybe by finding where $$w_d = w_0$$, which is the resonance frequency.

I made a graph. Can I consider steady-state amplitude as the driving force amplitude?

• Yes, you look for the resonance peak – FGSUZ Feb 7 at 19:35
• I'm not sure what you mean by resonance peak. – proxima Feb 7 at 20:28
• What is on the y axis of the graph? The displacement of the oscilaltor versus time, since the driving force starts to act? – nasu Feb 7 at 21:09
• @nasu The system is in equilibrium at x=0. – proxima Feb 7 at 21:21
• Driving force = input, steady state amplitude = output displacement. So no, driving force amplitude != steady state amplitude – oliver Feb 7 at 21:29